What does "conformally coupled scalar" mean?

Question

"Conformally coupled scalar $\phi$" - I encounter it a lot, but I can't find what it means.

Answer 1

A minimally coupled free scalar field is described by the action $$ S[g,\phi] = \frac{1}{2} \int d^D x \sqrt{g} g^{ab} \partial_a \phi \partial_b \phi . $$ However, this theory is not conformally invariant (i.e. invariant under Weyl transformations). In particular, if I rescale $g \to \Omega^2 g$ (where $\Omega$ is a function), then the action transforms as $$ S[\Omega^2 g,\phi] = \frac{1}{2} \int d^D x \Omega^{D-2} \sqrt{g}g^{ab} \partial_a \phi \partial_b \phi . $$ where the $\Omega^D$ factor comes from $\sqrt{g}$ and $\Omega^{-2}$ comes from $g^{ab}$. Clearly, the action is not conformally invariant (unless $D=2$). One way to fix this is to rescale $\phi$ as well by $\phi \to \Omega^{-\frac{1}{2}(D-2)} \phi$. We then find $$ S[\Omega^2 g,\Omega^{-\frac{1}{2}(D-2)} \phi] = \frac{1}{2} \int d^D x \sqrt{g} [ g^{ab} \partial_a \phi \partial_b \phi + {\cal O} \left( \partial \Omega \right) ] . $$ As is clear, the action is still not invariant under Weyl transformations because of terms proportional to the derivative of $\Omega$. To fix this, we add an extra term to the original action $$ S[g,\phi] = \frac{1}{2} \int d^D x \sqrt{g} \left( g^{ab} \partial_a \phi \partial_b \phi + \xi R[g] \phi^2 \right) . \tag{1} $$ It is then easy to show that for an appropriate choice of the constant $\xi$, this action is conformally invariant. A scalar field which couples to the background metric $g$ according to (1) is known as a "conformally coupled scalar" for obvious reasons.

Note that the action for a conformally coupled scalar reduces to the action of a minimally coupled scalar in flat spacetime. However, the two theories have different stress tensors.

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HW for the reader:

1. Show that the action (1) is conformally invariant iff. $\xi = \frac{D-2}{4(D-1)}$.

2. Find the stress tensor for the action (1) and show that it differs from the stress tensor of the minimally coupled scalar in flat spacetime. Verify that the stress tensor for the conformally coupled scalar is traceless but one for minimally coupled scalars is not (Tracelessness of the stress tensor is a requirement for Weyl invariance).